1)
For the step potential, show that for the region x > 0 and
E
< V o with K1 = Ö (2m(Vo - E) / h
The general solution of the appropriate Schrodinger equation is:
y(x) = C exp (K2 x) + D exp (-K2
x)
Show this.
Solution:
dy(x) / dx = CK2 exp (K2
x) – D K2 exp (-K2 x)
d2y(x) / dx2 =
CK22
exp (K2 x) + D K22 exp
(-K2 x)
=
K22 y(x) = y(x)
[2m(Vo - E) / h2]
Which
yields the appropriate Schrodinger eqn. on substitution:
-
h2/ 2m [2m(Vo - E) / h2]
y(x) + Vo y(x) = Ey(x)
2)
For the transmitted and reflected flux of the step potential, show that:
R + T = 1.
Solution:
We have: R = (K1 – K2)2 / (K1 + K2 )2
And:
T = 4K1K2 / (K1 + K2 )2
Then: R
+ T =
(K1 – K2)2 / (K1 + K2)2 +
4K1K2 / (K1 + K2 )2
= [K1 2 - 2K1K2 + K1 2 +
4K1K2] / (K1 + K2 )2
= [K1 2 +
2K1K2 + K1 2] / (K1 + K2 )2
= (K1 +
K2)2 / (K1 + K2 )2 =
1
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